Question

"Analogous to the Coulomb's law the Newton's law of gravitation states that whenever two bodies of certain mass $m_1$ and $m_2$ are separated by distance ${}^{\prime }{r}^{\prime }$, then each of them attract each other by a force given as:
$F=\frac{G{m}_1{m}_2}{r^2}$
where ${}^{\prime }{G}^{\prime }$ is a constant"
Now, two point charges $+q_1$ and $+q_2$ are placed on a horizontal table such that gravitational attraction between them balances electrostatic repulsion. What can be concluded about nature of equilibrium position ?

• A. It is a stable equilibrium position.
• B. It is an unstable equilibrium position.
• C. It is a neutral equilibrium position.
• D. The data is not sufficient to arrive at conclusion.

Answer 1

The correct option is C It is a neutral equilibrium position.


Let the mass of charge $q_1$ and $q_2$ is $m_1$ and $m_2$ respectively.
Considering charge $+q_2$,
$F_1$ is the electrostatic force of repulsion exerted by charge $q_1$ on $q_2$.
$F_2$ is the gravitational force of attraction exerted by mass of charge $q_1$ on $q_2$.
For equilibrium,
$N_2=m_{2}g$ (Normal reaction from table balances weight of charge $q_2$)
and $F_2=F_1$
$\Rightarrow \frac{G{m}_1{m}_2}{r^2}=\frac{k{q}_1{q}_2}{r^2}$
Thus, from the above relation it is concluded that charges will remain in equilibrium, irrespective of the distance between them.
i.e. $G{m}_1{m}_2=k{q}_1{q}_2$
$\Rightarrow$ The equilibrium position is therefore a neutral equilibrium.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: If equilibrium position is unaffected by displacement}\\ \text{of body, then it signifies the neutral equilibrium}\end{array}$